Lighting In Coaches

I am trying to install lighting into a Bachmann coach, with electricity fed from the track.

After getting the LED, I know I need to get a resistor too. Once the resistor and LED are soldered together, I need to complete the electrical circuit. If I feed and return onto the same point, it will short circuit. If I feed on one side of the bogie, and return on the other side ( so picking from one rail and returning on the other) that will short too, as the two rails will be linked. If I feed from and return on the same side of a bogie, but on different wheels, then I'm likely to create feedback.

Please can someone tell me, how I power my LED to complete a circuit.

Many Thanks Graham

Reply to
Graham Kendall
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But they'll be linked through a constant-voltage device (the LED, which drops *mumble* 0.7? *mumble* volts, and the resistor, which should be chosen such that its resistance gives a current of no more than 10mA or so (but check your LED spec first) from the maximum rail-to-rail voltage. Provided that your LED and resister haven't both gone open-circuit (most unlikely!) you won't be shorting the controller.

Eg if your controller puts out 12V, you want a resistance of (12-0.7)/0.01 = 1.1 K ohm or so.

R.

Reply to
Richard

Normally 0.6v, but close enough. Don't forget that 2 LEDs are needed, or the lighting will be directional.

Reply to
Matthew Sylvester

Matt,

Will I require a second resistor for the second LED, and I'm ssuming that I'm alright to connect the two LED's in Serie of each other?

Graham

Reply to
Graham Kendall

True

False.

No, this will create a short.

Right hand wheels --> resistor --> LEDs --> left hand wheels.

There will be no short, as the two rails are connected through a load.

You need two LEDs, wired in parallel and facing opposite directions. One will light when the train is running clockwise, and the other when the train is running counterclockwise. And LED is a Light Emitting DIODE, ie, it blocks current in one direction and passes it in the other, emitting light while doing so.

Hope this is clear enough. You ask a question that suggests you need a basic book on electrics for model railways. There are a number of them available.

HTH

Reply to
Wolf Kirchmeir

Thankyou Wolf that helps greatly. Would I not nead a second resistor too sit in your flow diagram above between the LEDs and the L/H Wheels to handle the current operating i nthe other direction

Yours

Graham Kendall

Reply to
Graham Kendall

Still wrong and more than a factor of two out ;-)

MBQ

Reply to
manatbandq

The LEDs are in parallel but one is reversed. The current goes through the resistor regardless of which LED is lit:

|---|>|---|

----| |----/\/\/\/------- |---|

Reply to
manatbandq

No.

Reply to
Wolf Kirchmeir

Yeah, quite right. I was thinking of a standard silicon PN junction.

Reply to
Matthew Sylvester

In message , " snipped-for-privacy@hotmail.com" writes

Surely the best solution is to go DCC?

Reply to
Jane Sullivan

wrote

Hide the capacitor(s) in the lavatory compartment (if corridor stock) or you can get loads of little ones (say 100mF) and make arrays that can hide under seat squabs, in a row in the roof lining, behind the battery boxes underneath, etc. The arrays will need to be in parallel (all the + wires ganged together, all the - wires ditto - a strip of veroboard will be ideal). Even so, this might only extend the LEDs' glowing life at a standstill to several seconds rather than instant darkness. Though does a spike-pulsed supply such as a Relco keep them continuously lit up? A small duty-cycle AC supply at high-frequency run through the rails might keep the lights glowing without turning the motor, which can't read the spikes because of the windings acting as an inductive filter. Someone will now say that as ever don't try this with a Portescap or other coreless because it'll turn into a door buzzer at a standstill...

A more ambitious plan - though I have not yet tried this - would be to rig up a supply to all the carriages using little polarised connectors under the buffer beams, and supply the LEDs from a PP3 rechargeable 9V battery (though a NiCd PP3 is actually 7.2V) concealed in the brake compartment of a typical coach rake or DMU set (I'm always assuming pre-1970s stock here and

4mm or bigger scale) and kept topped up by the bridge rectifier arrangement mentioned - or rather, if connected via a diode to the LEDs, the battery will cut out as soon as the supply reverse-biases the diode, and cut in again as a backup as the supply drops and lets the diode conduct again. The battery will need to be switchable via a concealed switch under the coach or disguised as a roof vent. A 7808 regulator (+8V) might have its uses here to stabilise the voltage from the track - you probably only need the little 100mA version. Maplin sells all the bits you need.

The annoying thing about LEDs is that the light colour is a bit unconvincing - yellows are way too yellow, while white ones give the cold light that suits fluorescent lights but can be toned with a bit of straw-coloured light gel or a dab of varnish with some yellow stirred in. Modern stock can use them as is for that "fridge with seats" look. White LEDs are quite cheap now - ALL Components, who come to quite a few large MREs sell a cooking version for about a pound each. Be careful not to run too much current through whites - they're a lot less tolerant about being overheated than the colour ones: bang too much in and they glow very bright white, then dull blue, then not at all, as the fluorescent coating on the junction roasts into powder - been there, done it recently, be warned!

Yellow LEDs look great tucked in the firebox doors of steam locos! And has anyone tried using white LEDs and a trigger/random switching circuit to recreate sparks from the shoebeams of third rail stock?

Tony Clarke

Reply to
Tony Clarke

Hi, Sorry to top post but you may find that using a bridge rectifier to be a better solution. Remember that with a bridge rectifier you will lose about 1.2 volts due to rectifier diode volts drop.

The volts drop across an LED is usually of the order of 2 volts, however the new white ones can drop about 3.2 volts. Normal forward current is about 15mA for a standard LED. Again the white ones normally handle a higher current due to their high intensity. Always check the data for an LED! Don't leave it to chance, especially with the whites LED costing a couple of quid or more.

The calculation of the current limiting resistor is as follows:

R = (Vsupply - Vled) / Idiode.

So for a supply of 12 volts (neglecting the PD across the bridge rectifier if you decide to install one) you would arrive at:

(12 - 2) / 0.015 = 666 ohms. Always aim on the high side of a resistor value, so use a 680R resistor. Power rating of said resistor is:

I * I * R

So the power in this case would be 153mW.

Hope that's clear now.

I also agree that DCC is the way to go because you have volts available at all times. Mind you because the DCC info is modulated at 100kHz you should use high speed rectifier diodes of you decide to use the rectifier option I mentioned. A smoothing cap and a small inductance would not go amiss either, after the rectifier I mean.

Cheers,

Rob.

Graham Kendall wrote:

Reply to
Robert Wilson

Right now you can buy your white LEDs in strings from the Christmas section at B&Q for between 10 and 15pence each. If you are doing lots of coaches they are a good deal. Might be even cheaper in the sale after Christmas, but of course might be sold out!

Keith

Reply to
Keith

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